Methanol (CH₃OH) is converted to bromomethane (CH₃Br) as follows: CH₃OH + HBr → CH₃Br + H₂O
If 12.23 g of bromomethane are produced when 5.00 g of methanol is reacted with excess HBr, what is the percentage yield?

Question

Quizplus · Accepted Answer

First, calculate the molar mass of methanol (CH3OH) and bromomethane (CH3Br). Methanol: 12.01 (C) + 4*1.01 (H) + 16.00 (O) = 32.05 g/mol. Bromomethane: 12.01 (C) + 4*1.01 (H) + 79.90 (Br) = 94.93 g/mol. Theoretical yield of bromomethane from 5.00 g of methanol: (5.00 g CH3OH / 32.05 g/mol) * 94.93 g/mol = 14.76 g. Percentage yield = (actual yield / theoretical yield) * 100 = (12.23 g / 14.76 g) * 100 = 82.9%. The closest answer provided is 82.6%.

Methanol (CH3OH) Is Converted to Bromomethane (CH3Br) as Follows: CH3OH

Methanol (CH₃OH) Is Converted to Bromomethane (CH₃Br) as Follows: CH₃OH